Moment of inertia of a rod with balls attached

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1. Aug 7, 2017

AfronPie

1. The problem statement, all variables and given/known data
Attached.

2. Relevant equations
I=mr^2, I=(1/12)m*l^2 for a rod.

3. The attempt at a solution
Part A I got by doing I=(1/12)3.5*(2.6)^2+2*.7*1.3^2 (I added the moment of inertia of the rod and the balls). Part B since the axis is on one of the balls, I thought we don't include that ball in the calculations. So the rod stays, (1/12)m*l^2 and we add the other .7 kg ball which is 2.6m away. So I_total=(1/12)3.5*2.6^2+.7(2.6)^2=6.7 kg*m^2. However that isn't the right answer. Can someone please explain to me where I went wrong. Any help is greatly appreciated.

2. Aug 7, 2017

3. Aug 7, 2017

AfronPie

Can you explain your hint. Why wouldn't you include the I of the rod?

4. Aug 7, 2017

TSny

You do need to include I for the rod. But when you move the axis of rotation to one end of the rod, does I for the rod remain the same?

5. Aug 7, 2017

AfronPie

Im confused. The only reason why I knew I=1/12ml^2 is because I looked up moment of inertia for a rod. The mass is the same and the length of the rod is the same? Can you please explain what it would change to and why?

6. Aug 7, 2017

TSny

Moment of inertia depends on how the mass of the object is distributed around the axis of rotation. When you switch the axis, you change how the mass is distributed relative to the new axis. Consult PAT (Parallel Axis Theorem).

http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html

7. Aug 7, 2017

AfronPie

Alright so I'm using the theorem. So I_total(rod)=I_cm+md^2=(1/12)3.5*2.6^2+4.9(1.3)^2=10.25
10.25+.7(2.6)^2=14.98 kg*m^2. Is this right?

8. Aug 7, 2017

TSny

It looks OK except for the factor of 4.9. How did you get that number?

9. Aug 7, 2017

AfronPie

It's the M in the parallel axis equation which I thought meant the sum of all the masses.

10. Aug 7, 2017

TSny

You should apply the theorem to just the rod. You are taking care of the point masses separately.

11. Aug 7, 2017

AfronPie

I_total(rod)=I_cm+md^2=(1/12)3.5*2.6^2+3.5(1.3)^2=7.89
7.89+.7(2.6)^2=12.62 kg*m^2. I just put it in and its right, thanks a lot for your help Tsny.

12. Aug 7, 2017

TSny

Great! Good work.